Question 1

The length of a wire is halved, and its radius is doubled. The resistance will change by a factor of

Accepted Answer

Resistance is inversely proportional to the cross-sectional area and directly proportional to the length. Halving the length reduces resistance by a factor of 1/2, and doubling the radius increases the area by a factor of 4, reducing resistance by another factor of 1/4. Combined, the resistance changes by a factor of 1/8.

Question 2

Consider the resistance of two wires made of the same material. Wire A has a length that is twice the length of wire B; the cross-sectional area of wire A is twice the cross-sectional area of wire B. The resistance of wire A is

Accepted Answer

The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area. Since wire A has twice the length (increasing resistance) and twice the cross-sectional area (decreasing resistance) compared to wire B, these changes cancel each other out, making the resistance of wire A the same as wire B.

Question 3

A 75-m length of copper wire whose resistivity is 1.7 $\times$ 10^-8 $\Omega$ . m has a resistance of 1.6 $\Omega$. The diameter of the wire is

Accepted Answer

To find the diameter, use the formula for resistance: $ R = \rho \frac{L}{A} $. Rearrange to find the cross-sectional area $ A = \frac{\rho L}{R} $. Then, use $ A = \pi (\frac{d}{2})^2 $ to solve for the diameter $ d $. Plugging in the values, $ A = \frac{1.7 \times 10^{-8} \times 75}{1.6} $, and solving for $ d $, we find $ d \approx 1.0 $ mm.

Question 4

The current through a copper wire (resistivity of 1.7 $\times$ 10^-8 $\Omega$ .m that has a radius of 1.5 mm and length of 100.0 m when a 1.5-V potential is applied across the ends of the wire is

Accepted Answer

The answer of The current through a copper wire (resistivity...

Question 5

A filament has a resistance of 0.10  $\Omega$at 20°C. If the resistance becomes 0.14  $\Omega$when the temperature of the filament increases to 90°C, the temperature coefficient of resistance for this material is

Accepted Answer

The temperature coefficient of resistance ($\alpha$) is calculated using the formula $\alpha = \frac{R_2 - R_1}{R_1 \times (T_2 - T_1)}$. Substituting the given values: $\alpha = \frac{0.14 - 0.10}{0.10 \times (90 - 20)} = 5.7 \times 10^{-3}/°C$.

Question 6

Negative temperature coefficients of resistivity

Accepted Answer

The answer of Negative temperature coefficients of resistivity
...

Question 7

A resistor of 1000  $\Omega$ has 150 mA flowing through it. The voltage across the resistor is

Accepted Answer

Ohm's law states that the voltage across a resistor is equal to the current flowing through it multiplied by the resistance of the resistor. In this case, the current is 150 mA, which is equal to 0.15 A, and the resistance is 1000 $\Omega$. Therefore, the voltage across the resistor is 0.15 A * 1000 $\Omega$ = 150 V.

Question 8

A hair dryer that draws 6.0 A when operating across a 110-V potential has a resistance of

Accepted Answer

The resistance of the hair dryer can be calculated using Ohm's Law, which is V = IR, where V is the voltage, I is the current, and R is the resistance. Rearranging the formula to solve for R gives R = V/I. Substituting the given values, R = 110V / 6.0A = 18.33 ohms, which is rounded to 18 ohms as per the options provided.

Question 9

Two resistors are connected in series, and the value of R₁ is twice the value of R₂. The resistor with the larger potential across it is

Accepted Answer

In a series circuit, the voltage drop across a resistor is directly proportional to its resistance. Since R₁ has twice the resistance of R₂, it will have a larger potential drop across it.

Question 10

Two resistors are connected in series, and the value of R₁ is twice the value of R₂. The resistor with the larger current through it is

Accepted Answer

In a series circuit, the current through each component is the same regardless of the resistance values.

Question 11

Two resistors are connected in parallel, and the value of R₁ is twice the value of R₂. The resistor with the larger potential across it is

Accepted Answer

In a parallel circuit, all components share the same voltage, regardless of their resistance values. Therefore, both resistors have the same potential across them.

Question 12

Two resistors are connected in parallel, and the value of R₁ is twice the value of R₂. The resistor with the larger current through it is

Accepted Answer

In a parallel circuit, the voltage across each resistor is the same. The current through a resistor is inversely proportional to its resistance (I = V/R). Therefore, the resistor with the lower resistance (R₂) will have a higher current through it.

Question 13

A 10- $\Omega$resistor is connected in series with a 30- $\Omega$ resistor. The total resistance of the combination is

Accepted Answer

When resistors are connected in series, their total resistance is the sum of their individual resistances. Therefore, 10 + 30 = 40 ohms.

Question 14

A 10- $\Omega$ resistor is connected in parallel with a 30- $\Omega$resistor. The total resistance of the combination is

Accepted Answer

The total resistance $R_{total}$ for resistors in parallel is given by $\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2}$. For a 10-ohm and a 30-ohm resistor in parallel, $\frac{1}{R_{total}} = \frac{1}{10} + \frac{1}{30} = \frac{3}{30} + \frac{1}{30} = \frac{4}{30} = \frac{1}{7.5}$, so $R_{total} = 7.5$ ohms.

Question 15

A 10.0-V potential is applied across two resistors, R₁ and R₂, connected in series. If the potential difference across R₁ is 4.0 V, the value of R₂ is

Accepted Answer

Since the total voltage is 10.0 V and the voltage across $R_{S1}$ is 4.0 V, the voltage across $R_{S2}$ must be $10.0 V - 4.0 V = 6.0 V$. In a series circuit, the voltage drop across a resistor is proportional to its resistance. Therefore, if $R_{S2}$ has a 6.0 V drop compared to $R_{S1}$'s 4.0 V drop, $R_{S2}$ must be $6.0 V / 4.0 V = 1.5$ times $R_{S1}$, making it 1.5 times the value of $R_{S1}$.

Question 16

When a 1.5-V potential is applied across a parallel network of two identical resistors, the net current is 1.0 A. The individual resistance of each resistor is

Accepted Answer

The total resistance of the parallel network can be found using Ohm's law, $V = IR$, where $V = 1.5$ V and $I = 1.0$ A, giving $R = \frac{V}{I} = \frac{1.5}{1.0} = 1.5 \Omega$. In a parallel network of two identical resistors, the total resistance is half of the resistance of one resistor, so each resistor has a resistance of $3.0 \Omega$.

Question 17

Five incandescent lightbulbs are connected in parallel. One of the bulbs burns out, causing the current through the other bulbs to

Accepted Answer

In a parallel circuit, each component is connected across the same two points. Each bulb has its own path to the power source. If one bulb burns out, the others are not affected because they still have a complete path to the power source. Therefore, the current through the other bulbs remains the same.

Question 18

Five incandescent lightbulbs are connected in series. One of the bulbs burns out, causing the current through the other bulbs to

Accepted Answer

When one bulb in a series circuit burns out, it breaks the circuit, causing the current to drop to zero. This is because a series circuit requires a complete path for current to flow, and a burned-out bulb interrupts this path.

Question 19

Two resistors, 20 $\Omega$ and 30  $\Omega$, are connected in parallel. A third resistor, 24  $\Omega$, is connected in series with the parallel combination. A 36-V power source is connected to the series-parallel combination. The current through the 24- $\Omega$ resistor is

Accepted Answer

First, calculate the equivalent resistance of the two parallel resistors: $R_{eq} = \frac{1}{\frac{1}{20} + \frac{1}{30}} = 12 \Omega$. Then, add the series resistor to find the total resistance: $R_{total} = R_{eq} + 24 = 36 \Omega$. Finally, use Ohm's law ($V = IR$) to find the current: $I = \frac{V}{R_{total}} = \frac{36V}{36\Omega} = 1.0 A$.

Question 20

Two resistors, 20  $\Omega$ and 30  $\Omega$, are connected in parallel. A third resistor, 24  $\Omega$, is connected in series with the parallel combination. A 36-V power source is connected to the series-parallel combination. The current through the 20- $\Omega$ resistor is

Accepted Answer

The equivalent resistance of the 20 Ω and 30 Ω resistors in parallel is 12 Ω. Adding the 24 Ω resistor in series gives a total resistance of 36 Ω. The total current from the 36-V source is 1 A. The current through the 20 Ω resistor is the same as the total current through the parallel combination, which is 0.60 A.

Quiz 27: Currents and Ohms Law

The length of a wire is halved, and its radius is doubled. The resistance will change by a factor of

Consider the resistance of two wires made of the same material. Wire A has a length that is twice the length of wire B; the cross-sectional area of wire A is twice the cross-sectional area of wire B. The resistance of wire A is

A 75-m length of copper wire whose resistivity is 1.7 $\times$ 10^-8 $\Omega$ . m has a resistance of 1.6 $\Omega$ . The diameter of the wire is

The current through a copper wire (resistivity of 1.7 $\times$ 10^-8 $\Omega$ .m that has a radius of 1.5 mm and length of 100.0 m when a 1.5-V potential is applied across the ends of the wire is

A filament has a resistance of 0.10 $\Omega$ at 20°C. If the resistance becomes 0.14 $\Omega$ when the temperature of the filament increases to 90°C, the temperature coefficient of resistance for this material is

Negative temperature coefficients of resistivity

A resistor of 1000 $\Omega$ has 150 mA flowing through it. The voltage across the resistor is

A hair dryer that draws 6.0 A when operating across a 110-V potential has a resistance of

Two resistors are connected in series, and the value of R₁ is twice the value of R₂. The resistor with the larger potential across it is

Two resistors are connected in series, and the value of R₁ is twice the value of R₂. The resistor with the larger current through it is

Two resistors are connected in parallel, and the value of R₁ is twice the value of R₂. The resistor with the larger potential across it is

Two resistors are connected in parallel, and the value of R₁ is twice the value of R₂. The resistor with the larger current through it is

A 10- $\Omega$ resistor is connected in series with a 30- $\Omega$ resistor. The total resistance of the combination is

A 10- $\Omega$ resistor is connected in parallel with a 30- $\Omega$ resistor. The total resistance of the combination is

A 10.0-V potential is applied across two resistors, R₁ and R₂, connected in series. If the potential difference across R₁ is 4.0 V, the value of R₂ is

When a 1.5-V potential is applied across a parallel network of two identical resistors, the net current is 1.0 A. The individual resistance of each resistor is

Five incandescent lightbulbs are connected in parallel. One of the bulbs burns out, causing the current through the other bulbs to

Five incandescent lightbulbs are connected in series. One of the bulbs burns out, causing the current through the other bulbs to

Two resistors, 20 $\Omega$ and 30 $\Omega$ , are connected in parallel. A third resistor, 24 $\Omega$ , is connected in series with the parallel combination. A 36-V power source is connected to the series-parallel combination. The current through the 24- $\Omega$ resistor is

Two resistors, 20 $\Omega$ and 30 $\Omega$ , are connected in parallel. A third resistor, 24 $\Omega$ , is connected in series with the parallel combination. A 36-V power source is connected to the series-parallel combination. The current through the 20- $\Omega$ resistor is

Quiz 27: Currents and Ohms Law

The length of a wire is halved, and its radius is doubled. The resistance will change by a factor of

Consider the resistance of two wires made of the same material. Wire A has a length that is twice the length of wire B; the cross-sectional area of wire A is twice the cross-sectional area of wire B. The resistance of wire A is

A 75-m length of copper wire whose resistivity is 1.7 ×\times× 10-8 Ω\OmegaΩ . m has a resistance of 1.6 Ω\OmegaΩ . The diameter of the wire is

The current through a copper wire (resistivity of 1.7 ×\times× 10-8 Ω\OmegaΩ .m that has a radius of 1.5 mm and length of 100.0 m when a 1.5-V potential is applied across the ends of the wire is

A filament has a resistance of 0.10 Ω\OmegaΩ at 20°C. If the resistance becomes 0.14 Ω\OmegaΩ when the temperature of the filament increases to 90°C, the temperature coefficient of resistance for this material is

Negative temperature coefficients of resistivity

A resistor of 1000 Ω\OmegaΩ has 150 mA flowing through it. The voltage across the resistor is

A hair dryer that draws 6.0 A when operating across a 110-V potential has a resistance of

Two resistors are connected in series, and the value of R1 is twice the value of R2. The resistor with the larger potential across it is

Two resistors are connected in series, and the value of R1 is twice the value of R2. The resistor with the larger current through it is

Two resistors are connected in parallel, and the value of R1 is twice the value of R2. The resistor with the larger potential across it is

Two resistors are connected in parallel, and the value of R1 is twice the value of R2. The resistor with the larger current through it is

A 10- Ω\OmegaΩ resistor is connected in series with a 30- Ω\OmegaΩ resistor. The total resistance of the combination is

A 10- Ω\OmegaΩ resistor is connected in parallel with a 30- Ω\OmegaΩ resistor. The total resistance of the combination is

A 10.0-V potential is applied across two resistors, R1 and R2, connected in series. If the potential difference across R1 is 4.0 V, the value of R2 is

When a 1.5-V potential is applied across a parallel network of two identical resistors, the net current is 1.0 A. The individual resistance of each resistor is

Five incandescent lightbulbs are connected in parallel. One of the bulbs burns out, causing the current through the other bulbs to

Five incandescent lightbulbs are connected in series. One of the bulbs burns out, causing the current through the other bulbs to

Two resistors, 20 Ω\OmegaΩ and 30 Ω\OmegaΩ , are connected in parallel. A third resistor, 24 Ω\OmegaΩ , is connected in series with the parallel combination. A 36-V power source is connected to the series-parallel combination. The current through the 24- Ω\OmegaΩ resistor is

Two resistors, 20 Ω\OmegaΩ and 30 Ω\OmegaΩ , are connected in parallel. A third resistor, 24 Ω\OmegaΩ , is connected in series with the parallel combination. A 36-V power source is connected to the series-parallel combination. The current through the 20- Ω\OmegaΩ resistor is

A 75-m length of copper wire whose resistivity is 1.7 $\times$ 10^-8 $\Omega$ . m has a resistance of 1.6 $\Omega$ . The diameter of the wire is

The current through a copper wire (resistivity of 1.7 $\times$ 10^-8 $\Omega$ .m that has a radius of 1.5 mm and length of 100.0 m when a 1.5-V potential is applied across the ends of the wire is

A filament has a resistance of 0.10 $\Omega$ at 20°C. If the resistance becomes 0.14 $\Omega$ when the temperature of the filament increases to 90°C, the temperature coefficient of resistance for this material is

A resistor of 1000 $\Omega$ has 150 mA flowing through it. The voltage across the resistor is

Two resistors are connected in series, and the value of R₁ is twice the value of R₂. The resistor with the larger potential across it is

Two resistors are connected in series, and the value of R₁ is twice the value of R₂. The resistor with the larger current through it is

Two resistors are connected in parallel, and the value of R₁ is twice the value of R₂. The resistor with the larger potential across it is

Two resistors are connected in parallel, and the value of R₁ is twice the value of R₂. The resistor with the larger current through it is

A 10- $\Omega$ resistor is connected in series with a 30- $\Omega$ resistor. The total resistance of the combination is

A 10- $\Omega$ resistor is connected in parallel with a 30- $\Omega$ resistor. The total resistance of the combination is

A 10.0-V potential is applied across two resistors, R₁ and R₂, connected in series. If the potential difference across R₁ is 4.0 V, the value of R₂ is

Two resistors, 20 $\Omega$ and 30 $\Omega$ , are connected in parallel. A third resistor, 24 $\Omega$ , is connected in series with the parallel combination. A 36-V power source is connected to the series-parallel combination. The current through the 24- $\Omega$ resistor is

Two resistors, 20 $\Omega$ and 30 $\Omega$ , are connected in parallel. A third resistor, 24 $\Omega$ , is connected in series with the parallel combination. A 36-V power source is connected to the series-parallel combination. The current through the 20- $\Omega$ resistor is